Problem: Find $\lim_{x\to \infty}(3+3e^x)^{^{\frac{3}{x}}}$. Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $3$ (Choice C) C $e^3$ (Choice D) D The limit doesn't exist.
Taking $x$ to $\infty$ in $(3+3e^x)^{^{\frac{3}{x}}}$ results in the indeterminate form $\infty^{^{0}}$. To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting $y=(3+3e^x)^{^{\frac{3}{x}}}$, we will find $\lim_{x\to \infty}\ln(y)$. Once we find it, we will be able to find $\lim_{x\to \infty}y$. $\ln(y) =\dfrac{3\ln(3+3e^x)}{x}$ Taking $x$ to $\infty$ in $\dfrac{3\ln(3+3e^x)}{x}$ results in the indeterminate form $\dfrac{\infty}{\infty}$, so now it's L'Hôpital's rule's turn to help us with our quest! $\begin{aligned} &\phantom{=}\lim_{x\to \infty}\ln(y) \\\\ &=\lim_{x\to \infty}\dfrac{3\ln(3+3e^x)}{x} \\\\ &=\lim_{x\to \infty}\dfrac{\dfrac{d}{dx}[3\ln(3+3e^x)]}{\dfrac{d}{dx}[x]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to \infty}\dfrac{\left(\dfrac{9e^x}{3+3e^x}\right)}{1} \\\\ &=\lim_{x\to \infty}\dfrac{9e^x}{3+3e^x} \\\\ &=3 \gray{\text{The leading terms are the coefficients of $e^x$}} \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to \infty}\dfrac{\dfrac{d}{dx}[3\ln(3+3e^x)]}{\dfrac{d}{dx}[x]}$ actually exists. We found that $\lim_{x\to \infty}\ln(y)=3$, which means $\lim_{x\to \infty}y=e^3$. [Why?] In conclusion, $\lim_{x\to \infty}(3+3e^x)^{^{\frac{3}{x}}}=e^3$.